STRUCTURE OF RHODIUM ISOTOPES
By Prof. L, Kaliambos (Natural Philosopher in New Energy) September 2014 In 1964 Gell--Mann and Zweig discovered the charged quarks responsible for revealing the charge distributions in nucleons able to give the electromagnetic strong force for the correct nuclear structure. However the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favor of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons Here among 288 quarks one sees 9 charged quarks in proton and 12 ones in neutron able to give the charge distributions in nucleons for the nuclear binding and nuclear structure by applying the laws of electromagnetism (See my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS ). Naturally occurring rhodium (Rh) is composed of only one stable isotope, Rh-103. The most stable radioisotopes are Rh-101 with a half-life of 3.3 years, Rh-102 with a half-life of 207 days, Rh-102m with a half-life of 2.9 years, and Rh-99 with a half-life of 16.1 days. Thirty other radioisotopes have been characterized with atomic weights ranging from 88.949 u (Rh-89) to 121.943 u (Rh-122). Most of these have half-lifes that are less than an hour except Rh-100 (half-life: 20.8 hours) and Rh-105 (half-life: 35.36 hours). There are also numerous meta states with the most stable being Rh-102m (0.141 MeV) with a half-life of about 207 days and Rh-101m (0.157 MeV) with a half-life of 4.34 days. The primary decay mode before the only stable isotope, Rh-103, is electron capture and the primary mode after is beta emission. The primary decay product before Rh-103 is ruthenium and the primary product after is palladium. ' ' STRUCTURE OF Rh-99, Rh-101, AND Rh-103 WITH S = -1/2 ''' For understanding the structure of the above nuclides you must read my STRUCTURE OF Rh-103 . The structure of the above nuclides with odd number of extra neutrons is based on the structure of Rh-90 with 45 protons and 45 neutrons. In the following diagram of the Rh-90 we clear that the structure of Rh-90 is based on the structure of Ru-88 with S =0 in which the one additional vertical system of p45n45 with S= 0 breaks the high symmetry of Ru-88. '''DIAGRAM OF Rh-90 WITH S =0 In this structure you see the additional vertical system of p45n45 with S =0 which makes horizontal bonds with the n25 and p27 of the structure of Ru-88 with S=0 . Note that in the Ru-88 the p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' ' ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12........n12......p32' ' -HP6 p31....n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' n27.........p8..........n8.......p28' ' -HP4 n45....p27.....n7..........p7.......n28 ' ' p25.........n6.........p6.......n26' ' +HP3 p45...n25……p5........n5……...p26 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2………p2........n22' ' +HP1 n21......p1........n1........p22 ' ' p37......n37 ' ' -HSQ n39......p39 ' However for getting the only one stable Rh-103 with S = -1/2 after a careful analysis we found that the above structure of Rh-90 with S=0 in the presence of an odd number of extra neutrons changes the structure . Here for symmetrical arrangements the n40p40 changes the spin from S = +1 to S =0 giving S = -1, because the p40 and n40 move from +HQS to p26 and n28 in order to make the symmetrical horizontal bonds . Under this condition the Rh-90 has S = -1. Then in the presence of 5 extra neutrons of positive spins and 4 extra neutrons of negative spins we get the structure of Rh-99 with S = -1/2. That is S = -1 +5(+1/2) + 4(-1/2) = -1/2. Then adding two extra neutrons of opposite spins we get the structure of Rh-101 with the same S = -1/2. Note that Rh-99 and Rh-101 are unstable nuclides because the small number of extra neutrons cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the presence of two more extra neutrons than those of Rh-101 we get the stable Rh-103 with S =-1/2, because the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. ' ' STRUCTURE OF Rh-100, AND Rh-102 WITH S =-1 Similarly the structure of the above unstable nuclides with even number of extra neutrons is based on the structure of Rh-90 with S = -1. For example the Rh-102 with S = -1 has 12 extra neutrons of opposite spins . ' ' STRUCTURE OF Rh-104, Rh-106, Rh-108, Rh-112, Rh-114, AND Rh-116 WITH S = +1 ' In this group of the above unstable nuclides we see that the structures are based on a new structure of Rh-90 with S = +1. In this case for symmetrical arrangements the p37n37 changes the spin from S = -1 to S = 0 giving S = +1. Particularly as in the case of Rh-90 the n37 and p37 move from the -HSQ to the p26 and p26 for making symmetrical horizontal bonds. Then in the presence of extra neutrons one gets the structures of the above nuclides. For example the Rh-116 with S =+1 has 26 extra neutrons of opposite spins. ' ''' '''STRUCTURE OF Rh-94, Rh-98, Rh-110, AND Rh-118 After a careful analysis I found that in the presence of such even number of extra neutrons the structures of the above unstable nuclides are based on another structure of Rh-90 with S = +2. In this case the p37n37 changes the spin from S = -1 to S =+1 giving S = +2. Particularly it moves from the -HSQ to +HSQ in order to make horizontal bonds with n40p40. Then adding extra neutrons we get the structures of the above nuclides. For example the Rh-118 with S = +4 has 4 extra neutrons of positive spins and 24 extra neutrons of opposite spins giving S = 0. That is S = +2 + 4(+1/2) + 0 = +4 . ' ' STRUCTURE OF Rh-91, Rh-93, Rh-95, Rh-97, Rh-109, Rh-111, Rh-113, Rh-115, Rh-117, Rh-119, AND Rh-121 Here we see that the structures of the above unstable nuclides are based on the structure of Rh-91with S = +7/2 . Especially the structure Rh-91 is based on a new structure of Rh-90 with S =+ 4 . In this case the p31937 and p39n39 change their spins from S =-2 to S = +2 giving S =+4. Particularly they move from the -HSQ to +HSQ in order to make horizontal bonds with p38n38 and n40p40. Son in the presence of the one extra n46(-1/2) the structure of Rh-91 has S = +7/2 . That is S = +4 + 1(-1/2) = +7/2 Then adding extra neutrons we get the structures of the above nuclides. For example the Rh-121 with S = + 7/2 has 30 more extra neutrons with opposite spins than the one n46. ' ' NUCLEAR STRUCTURE OF Rh-89 WITH S = +7/2 Similarly in the absence of two neutrons of opposite spins in the structure of Rh-91 with S = +7/2 we get the structure of Rh-89 with the same S = +7/2 . Category:Fundamental physics concepts